Double integral -II

[canvas]

Thank you guys, I got it now, was so easy but I'm blind sometimes...
What about next integral, integration area is so weird...
[math]\iint_D(x+y)dxdy,\,\,D:\,\,x\geq0,\,\,y\geq0,\,\,\sqrt{x}+\sqrt{y}\leq1[/math]

I'm talking about this integral, it seems like you re talking about integral from the main post

 

[nasi112]

Do you mean \(\displaystyle \sqrt{x} - \sqrt{y} \leq 1\)?

Or maybe \(\displaystyle \sqrt{x} + \sqrt{y} \geq 1\)

 

[canvas]

Yeah I mean sqrt(x)+sqrt(y)<=1

 

[nasi112]

Since this is a new problem, can you please post it in a new thread?

 

[canvas]

Since this is a new problem, can you please post it in a new thread?

OK

 

[canvas]

Someone created a new post from the previous one, thank you administration

 

[nasi112]

Ok. Now, if you are working in the real analysis, this inequality is wrong,

\(\displaystyle \sqrt{x} + \sqrt{y} \leq 1\)

Why?

Assume that \(\displaystyle x = 4\), then

\(\displaystyle \sqrt{4} + \sqrt{y} \leq 1\)

\(\displaystyle 2 + \sqrt{y} \leq 1\)

\(\displaystyle \sqrt{y} \leq -1\)

Can you give me a number if I plug it in \(\displaystyle \sqrt{y}\), it will give me \(\displaystyle -1\) or less?

 

[nasi112]

Ok. Now, if you are working in the real analysis, this inequality is wrong,

\(\displaystyle \sqrt{x} + \sqrt{y} \leq 1\)

Why?

Assume that \(\displaystyle x = 4\), then

\(\displaystyle \sqrt{4} + \sqrt{y} \leq 1\)

\(\displaystyle 2 + \sqrt{y} \leq 1\)

\(\displaystyle \sqrt{y} \leq -1\)

Can you give me a number if I plug it in \(\displaystyle \sqrt{y}\), it will give me \(\displaystyle -1\) or less?

Sorry, it is my bad. I was working on \(\displaystyle \sqrt{x} + \sqrt{y} \leq 0\)

 

[nasi112]

My question for you is, are you challenging yourself to move this integral from cartesian coordinate to polar coordinate, or the instructor is asking you to do that?

Usually, we change the integral to polar to make calculations easier. This integral can be solved easily in cartesian coordinate.

 

[canvas]

I'm trying to solve it both ways: in Cartesian coordinate system, and polar coordinate system. But I don't have any idea for any, I'm just a beginner in double integrals, can you give me a hint how to solve it in Cartesian coordinate? What's the range of integration for x and y?

 

[BigBeachBanana]

I'm trying to solve it both ways: in Cartesian coordinate system, and polar coordinate system. But I don't have any idea for any, I'm just a beginner in double integrals, can you give me a hint how to solve it in Cartesian coordinate? What's the range of integration for x and y?

Graphs are your friends. They help you determine the limits. There are two ways to solve the problem: dxdy or dydx. My personal preference is dydx. By holding x constant from 0 to 1, y varies between 0 and the curve [imath]\sqrt{x}+\sqrt{y}\le1 \implies y\le (1-\sqrt{x})^2[/imath]

 

[canvas]

Graphs are your friends. They help you determine the limits. There are two ways to solve the problem: dxdy or dydx. My personal preference is dydx. By holding x constant from 0 to 1, y varies between 0 and the curve [imath]\sqrt{x}+\sqrt{y}\le1 \implies y\le (1-\sqrt{x})^2[/imath]
View attachment 30785

I was afraid about squaring both sides, because of the sign of RHS

 

[BigBeachBanana]

I was afraid about squaring both sides, because of the sign of RHS

This is the graph of [imath]y= (1-\sqrt{x})^2[/imath]. Although it shows more than what we're originally wanted, but when you integrate w.r.t x, it restricts from 0 to 1.

 

[nasi112]

You will not be able to solve in polar coordinate because you cannot isolate \(\displaystyle r\).

In the cartesian coordinate, it will be,

\(\displaystyle \int_0^1 \int_0^{(1-\sqrt{y})^2} (x + y) \ dx \ dy\)

 

[canvas]

[math]\int_0^1(\int_0^{x-2\sqrt{x}+1}(x+y)dy)dx\\\\\text{is it okay?}[/math]

 

[nasi112]

[math]\int_0^1(\int_0^{x-2\sqrt{x}+1}(x+y)dy)dx\\\\\text{is it okay?}[/math]

If you want the \(\displaystyle dy\) comes before \(\displaystyle dx\), it is ok too.

\(\displaystyle \int_0^1 \int_0^{(1-\sqrt{x})^2} (x + y) \ dy \ dx\)

 

[canvas]

This is the graph of [imath]y= (1-\sqrt{x})^2[/imath]. Although it shows more than what we're originally wanted, but when you integrate w.r.t x, it restricts from 0 to 1.


View attachment 30787

why don'w we take the area from (1,+inf) ?

 

[BigBeachBanana]

why don'w we take the area from (1,+inf) ?

I'm assuming you're referring to x's limits. Look at the original region in post #11. X's limits only range from 0-1. The addition region 1 to infinity was introduced because we squared y. But if you were to take from 1-infinity, the definite integral diverges because there will be an infinite amount of area.

 

[canvas]

"X's limits only range from 0-1."

Can't see that ;/

 

[BigBeachBanana]

"X's limits only range from 0-1."

Can't see that ;/

Let's look at the inequality analytically: [math]0 \le \sqrt{x}+\sqrt{y} \le 1[/math]Since it is bounded by 1, and you can't have negatives inside square roots, then the least and most x or y can be is between 0 and 1. Does this help?